A good day to everyone.

Consider the following "Conjecture":

If $a, b \in M \subset \mathbb{N}$, then $1 < a < b$ and ... [plus some more conditions on $a, b$ and $M$...] if and only if $\sigma(a) < \sigma(b)$.

I have two questions at this point:

(1) What properties should elements of the set $M$ have to satisfy this "Conjecture"?

(2) What happens to the sets in $M$ if you restrict to the case $\gcd(a, b) = 1$? (Of course, the sets in (1) should have greater asymptotic density than the sets in (2), but can this notion be made more precise?)

Now, for the motivation: An example of a set $M$ satisfying this "Conjecture" is any pair $(A, B)$ of relatively prime factors of an odd perfect number given in the Eulerian form $N = {q^k}{n^2}$ (modulo some exceptions), where $\gcd(q, n) = 1$ and $q \equiv k \equiv 1 \pmod4$.

Let $$I(x) = \displaystyle\frac{\sigma(x)}{x}$$ be the abundancy index of $x \in \mathbb{N}$. Then by enumerating all possible permutations of the set

$$q^k, n, \sigma(q^k), \sigma(n) \in \mathbb{N}$$

we know that exactly one of the following holds (because of $1 < I(q^k) < I(n)$):

$$[1] \hspace{0.2in} q^k < n < \sigma(q^k) < \sigma(n)$$ $$[2] \hspace{0.2in} q^k < \sigma(q^k) < n < \sigma(n)$$ $$[3] \hspace{0.2in} n < q^k < \sigma(q^k) < \sigma(n)$$ $$[4] \hspace{0.2in} n < q^k < \sigma(n) < \sigma(q^k)$$ $$[5] \hspace{0.2in} n < \sigma(n) < q^k < \sigma(q^k)$$

Last question: Which of these five "configurations" could we eliminate?